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Proving a function is well defined and continuous

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f_{n}(x)=\frac{-x^2+2x-2x/n+n-1+2/n-1/n^2}{(n ln(n))^2}[/itex]

    Prove [itex]f(x) = \sum^{\infty}_{n=1} f_{n}(x)[/itex] is well defined and continuous on the interval [0,1].

    2. Relevant equations

    In a complete normed space, if [itex]\sum x_{k}[/itex]converges absolutely, then it converges.


    3. The attempt at a solution

    Working in a complete normed space [itex](C[0,1], || . ||_{∞})[/itex],

    consider the real series [itex]\sum^{∞}_{n=1}||f_{n}||_{∞}=\sum^{∞}_{n=1} sup <f_{n}(x) : x\in[0,1]> [/itex]

    It just remains to show that [itex]\sum^{∞}_{n=1}|f_{n}|[/itex] converges, but I can't seem to figure out how. Could anyone help me out here?
     
    Last edited: Feb 20, 2012
  2. jcsd
  3. Feb 20, 2012 #2

    lanedance

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    some ideas i haven't tried them yet though... how about first separating n into some manageable pieces...

    first evaluate the convergence of the terms only dependent on n, then consider the terms with an x and have a think about what x value will maximize that sum.

    then if it's tough to pull it together you may want to consider some properties of the norm.

    then I think you may need more for the continuous part and move to the epsilon deltas
     
  4. Feb 20, 2012 #3
    Rather than dealing with all the terms separately, I've been trying to just apply a simple comparison test (followed by an integral test) to prove its convergence, but I could only come up with divergent cases.

    I'm starting to think if there is an error with this question; does this series converge if it starts at n=1?
     
  5. Feb 21, 2012 #4

    lanedance

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    yeah as 1/ln(1)^2 is undefined
     
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